# Solving Systems of Equations by Graphing Review | Expressions & Equations | Grade 8

Now we would like to graph the following system and find the point of intersection. In our first system we have y equals x, plus five and y equals one half x, plus four. So when i see these systems it’s very important that they are in the slope, intercept form or the y equals inner y equals mx plus b form. So let’s go ahead and write each one of these equations out and then i will we’ll break up c find out. Where is our intercept and where is our slope y equals one half x plus four? So we know right away that, because it’s in the y equals mx plus b b is going to be our intercept. Έτσι 5 is going to this. Is our y intercept right here and then our slope is the number in front of x and we don’t see a number so it’s, our imaginary one, and that is represented by the letter of m, and this is going to be our slope. So the slope is the rise over run and because it’s, a one one over one, is still one and our intercept is going to be five. So let’s go ahead and plug this in and i’m going to use red for this. First um line and i’ll use blue for my second line, so here we are. Our starting point is on the y axis at 0 5, and here it is right here at 5 and then our slope is going to be up 1 πάνω από 1.

plot. The next point up 1 πάνω από 1 plot, the next point up one over one. I can go ahead and pull out my line and see how it matches up right here, let’s get it all nice and straight very good. Τώρα, our second linear function. Our second equation, is we see that the b or the y intercept is going to be four, so we are going to look for four on the y axis and the four is right here and then we will see our slope. Our slope, represented by the letter of m, our slope, is already in a fraction form, and so that tells us that we are going to go up 1 and to the right two. So here we start at our point and we will say: we’ll go up one over one, two up one over one, two and you’re saying maybe say to yourself. This is not necessarily going to intersect, but it is so it’s going to intersect right there and it’s. If i went back one, i went down one in a negative and back two. It would still be a negative one over a negative two is a positive one. Half so our point of intersection is right here now. What is that final solution? What is that point of intersection? So our solution is going to be an ordered pair and that will go from the origin. We’Ll go our x on our x axis negative two, and then we will go up one two, three up three on our y axis, Εντάξει, and so our solution is negative.

Two three let’s go ahead and try it once more. Here we go. Let’S put our let’s erase all of this and let’s put all our lines back. We will be using these lines again so i’m, just going to put them right here off to the side, and here we go so we have. Our next equation is going to be y equals negative x, minus two and two x minus y equals two. So in this sense, let’s write our equations out again y equals negative x minus two and then 2x minus y equals 2. i’m going to color code them. Just so we know where it what lines we will be using for each equation to represent each equation, and so we see right away that this negative 2 is already our y intercept, and so we are going to start at a negative 2 right here at a Negative 2, our slope. Now we see that our slope is that imaginary one, because there was no coefficient that we could see so it’s, really a negative one over one. So that means we are going to go down one and over to the right one. So let’s go down one over one down one over one down one over one: let’s go ahead and plot our red line; let’s turn it around and we’ll pull it right back up. Here we go great and extend it. We want to extend it as far as we possibly can on the graph to make sure that we don’t miss any point of intersection now.

The second equation is not in the y equals mx, plus b form or the point slope form, not the point slope formula, but the slope intercept formula and so we’re going to need to isolate y and so i’m going to add y to both sides. And that is going to give me 2x equals y plus 2. Then we are going to subtract the 2 because we need to get the y by itself. So we’re going to subtract 2 on both sides, giving me y equals 2x minus 2.. So when we go from here, we see that that negative 2 is going to be our starting point, and so we are going to start at a negative two and then we are going to in our slope. We see that it’s negative two over one. So we are going to rise to over one rise, two over one rise, two over one. It kind of was pretty simple, because our graph – Εδώ, i should say, was the point of intersection – was right there at our there. You go right there at the intercept, so we could see that our point of intersection is right here in front of zero. So our solution remember our solution is always an ordered pair. Our solution is zero. Negative two started here at the origin. We don’t move left or right it’s zero, and then we go down the y axis to a negative two and that’s our solution.

So i hope you get a little bit more practice in reviewing solving systems by graphing. Make sure you hit the bell notifications and subscribe to our channel, so you know the next time a video of ours come out so and don’t forget to hit that like as well.